Question

Two parallel line l and m are intersected by a transversal 'p' show that the quadrilateral formed by bisectors of interior angel is a rectangle.

Answer 1

Given : $l\parallel m$

Transversal p intersects l & m at A & C respectively. Bisector of $\angle$ PAC & $\angle$ QCA meet at B. And, bisector of $\angle$ SAC & $\angle$ RCA meet at D.

To prove : ABCD is a rectangle.

Proof :

We know that a rectangle is a parallelogram with one angle ${90}^o$.

For $l\parallel m$ and transversal p

$\angle PAC=\angle ACR$

So, $\frac{1}{2}\angle PAC=\frac{1}{2}\angle ACR$

So, $\angle BAC=\angle ACD$

For lines AB and DC with AC as transversal $\angle BAC$ & $\angle ACD$ are alternate angles, and they are equal.

So, $AB\parallel DC$.

Similarly, for lines BC & AD, with AC as transversal $\angle BAC$ & $\angle ACD$ are alternate angles, and they are equal.

So, $BC\parallel AD$.

Now, In ABCD,

$AB\parallel DC$ & $BC\parallel AD$

As both pair of opposite sides are parallel, ABCD is a parallelogram.

Also, for line $l$,

$\angle PAC+\angle CAS={180}^o$

$\frac{1}{2}\angle PAC+\frac{1}{2}\angle CAS={90}^o$

$\angle BAC+\angle CAD={90}^o$

$\angle BAD={90}^o$.

So, ABCD is a parallelogram in which one angle is 90${}^o$.

Hence, ABCD is a rectangle.